a(n)³ + a(n)² * [1 - a(n + 1)] + 1= a(n + 1)
-> a(n)³ + a(n)² + 1= a(n + 1) * [1 + a(n)²]
-> a(n + 1) = [a(n)³ + a(n)² + 1] / [a(n)² + 1] = a(n)³ / [a(n)² + 1] + 1
又a1 = 1,所以 an ≥ 1
a(n + 1) / an = [a(n)³ + a(n)² + 1] / [a(n)³ + an]
an ≥ 1 -> a(n)² ≥ an -> a(n)² + 1 > an -> a(n)³ + a(n)² + 1 > a(n)³ + an
-> a(n + 1) / an > 1
-> a(n + 1) > an